Archive for April, 2015

Problem 672

Consider the function F and G in the graph.

Find (F – G)(3)

gp

Solution:-

Being by applying the definition for the difference of two function values to rewrite(F -G)(3)as F(3) – G(3).The difference is graphically equivalent to the difference of the second coordinates on F and G when the first coordinate is 3.

To find the difference of the function values F(3)and G(3), evolution involves a three step process.

1. Find the value of F(3)

2. Find the value of G(3)

3. Then find the difference of F(3)and G(3)

The first step is to find the value of the  F(3), which is the y-coordinate of the point on the graph F wneh the x-coordinate is 3.

F(3) = 1

The second step is to find the value of G(3), which is the y-coordinate of the point on the graph G when the x-coordinate is 3.

G(3) = 0

The third step is to find the difference of F(3) and G(3).

(F – G)(3) = F(3)- G(3) = 1 – 0 =1

 

 

 

Problem 671

For the pair of function f and g, determine the domain of f/g.

F(x) = \frac{2}{x-3},g(x) = 4-x

 

Solution:-

 

The domain of t/g is the set of all values common to the domains of t and g, excluding values for which g(x) is 0.

The domain of a function defined by an equation is the set of all number for which real values of the function can be calculated. A function can be undefined at a number because calculating its value results in an undefined in an undefined operation like division by zero or an even root of a negative number.

For the function , f(x) =\frac{2}{x-3}. The denominator is zero for x= 3, which result in an undefined value. Therefore, the domain of f is {x I x is  a real number and x\neq3}.

For the function , g(x) = 4 – x, a value can be calculated for any real number x. Therefore, the domain of g is {x I x is a real number}.

The domain of the quotient also excludes all values for which f(x) is zero. Therefore solve the equation

4 – x = 0.

X = 4

The domain of f/g is the set of all values common to { x I x is a real number and x \neq3} and {x I x is a real number}, and also excluding x = 4.

Therefore , the domain of f/g is {x I x is a real number and x \neq 3 and  x \neq4}.

 

Problem 670

Using the graph on the right, find the value of the following.

(f-g)(1)

graph

 

Solution:-

If f(x) represent one function, g(x) represent a second function, and x is in the domain of both function, then

(g – f) (x) = g(x) – f(x).

The procedure to find g(1) – f(1) involves three steps.

1. Evaluate g(1).

2. Evaluate f(1).

3. Subtract the value, g(1) – f(1).

 

 

1. To find the value, g(1), use the graph.

g(1) = -1

 

2. To find the value, f(1), use the graph.

f(1) = 2

 

3. Subtract the values to find the final answer.

(g – f) (1) = -3

 

Problem 669

Kara’s Custom Tees experienced fixed cost of 400 and variable cost of 5 a shirt. Write an equation that can be used to determine the total expenses encountered by Kara’s Custom Tees. Let x be the number  of shirt, and let C(x) be the total cost of producing x shirt. Then, calculate the cost of producing 8 shirts.

 

Solution:-

 

Notice that the cost of producing zero shirt is the fixed 400. Write the point (0,400)as a solution to the equation.

If 10 shirts are produced, the total cost 450.

Write a second point(10,450) as a solution to the equation.

Next , find the slope of the line through the point, (0,400) and (10,450).

m  = \frac{450 - 400}{10} = 5

Since  (0,400) is the y-intercept and 5 is the slope, use the formula y = mx + b

to write the equation. In this case, y = C(x), m = 5, and b = 400.

Thus, C(x) = 5x + 400.

By plotting the two point, you get graph of C(x) = 5x + 400 with domain [0,\infty).

Calculate the cost of  producing 8 shirts. In other world, find C(8).

C(8) = 5(8) + 400

C(8) = 40 +400

C(8) = 440

The cost of producing 8 shirt is 440.

 

Problem 668

Media Services charges $40 for a phone and $20/month for its economy plan. Find a linear model that determines the total cost, C(t), of operating a Media Services phone for t months.

Use the model to find the total cost for 4 months of service.

 

Solution:-

 

Formulate a linear model that can be used to determine the total cost, C(t), of operating a Media Services phone for t months.

The total cost involves the initial purchase of a phone plus a monthly fee of %20. The initial purchase cost of %40 is a one-time charge which is a constant term in the model.

The monthly fee is a variable cost based on the number of months of months of service. Remember that t represents the number of months of service for a phone.

The model equation would be C(t) = 20t + 40.

To find the total cost for 4 months of service, substitute 4 for t and evaluate c(4).

C(4) = 20(4) + 40

C(4) = 120

Therefore, the cost of operating a  phone for 4 months is $120.

Problem 667

If (x,y) is a point on the unit circle in quadrant IV and if x = \frac{1}{2}, what is y?

 

Solution:-

y = -  \frac{\sqrt{3}}{2}

Problem 666

If (x,y) is a point on the unit circle in quadrant IV and if x = \frac{\sqrt{2}}{2}, what is y?

 

Solution:-

 

If the point (x,y) is on the unit circle, then the point satisfies the equation of the unit circle

x^{2}+y^{2}=1

(\frac{\sqrt{2}}{2})^{2}+y^{2}=1

Evaluate (\frac{\sqrt{2}}{2})^{2}

 

\frac{1}{2}+ y^{2} = 1

 

Solve for y.

y^{2} =  \frac{1}{2}

 

Take the square root of both sides.

y = \pm \sqrt{\frac{1}{2}}

y = \pm  \frac{\sqrt{2}}{2}

 

In quadrant IV, the y-coordinate of the point is negative. Therefore, y = – \frac{\sqrt{2}}{2}

 

 

Problem 665

What point is symmetric with respect to the y-axis to the point(  \frac{\sqrt{3}}{2} ,\frac{-1}{2})?

 

Solution:-

 

Recall  for any point (x,y), the point symmetric about the y-axis to (x,y) is the point (-x,y).

 

If two point are symmetric about the y-axis, then the point have the same y-coordinate and opposite x-coordinates. Replace the x-coordinate of the given point with its opposite.

 

Identify the x-coordinate of the given point(  \frac{\sqrt{3}}{2} ,\frac{-1}{2}).

 

X = \frac{\sqrt{3}}{2}

If x = \frac{\sqrt{3}}{2}, than the opposite of x, -x, = –\frac{\sqrt{3}}{2}.

Therefore , the point symmetric to the point (  \frac{\sqrt{3}}{2} ,\frac{-1}{2}) with respect to the y-axis is (  \frac{-\sqrt{3}}{2} ,\frac{-1}{2})

 

Problem 664

Find the length s and area A.

 

area

 

Solution:-

 

The length of the arc is 4.538 yards.

 

The area A of the sector is 29.496 square yards.

 

Problem 663

Find the central angle \theta which forms a sector of  area 21 square feet of a circle of radius 11 feet.

 

Solution:-

A central angle is an angle whose vertex is at the center of a circle. The rays of a central angle subtend an arc on the circle. The part of the circle between the rays of the angle and the arc subtended is called a sector.

The area A of a sector of a circle of radius r formed by a central angle of  \theta radians is equal to the following.

 

 

A = \frac{1}{2}r^{2}\theta

 

Solving for \theta=\frac{2A}{r^{2}}. Substitute the values for A and r, and simplify.

\theta=\frac{2A}{r^{2}}

=\frac{2*21 square feet}{(11 feet)^{2}}

\approx 0.347 radians

Therefore , the central angle \theta which from a sector a sector of area 21 square feet of a circle of radius 11 feet is \theta = 0.347 radians.